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Home » To choose exactly 2 papers by Newton and 1 by Leibniz: - AMAZONAWS

To choose exactly 2 papers by Newton and 1 by Leibniz: - AMAZONAWS

📅 March 5, 2026 👤 scraface

Mar 05, 2026

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📌 Question: Find the quadratic polynomial $ p(x) $ such that $ p(1) = 4 $, $ p(2) = 11 $, and $ p(3) = 22 $.

📌 Solution: Let $ p(x) = ax^2 + bx + c $. Using the given values:

📌 a(1)^2 + b(1) + c = a + b + c = 4, \quad a(4) + 2b + c = 4a + 2b + c = 11, \quad 9a + 3b + c = 22.

📌 Subtract the first equation from the second: $ 3a + b = 7 $. Subtract the second from the third: $ 5a + b = 11 $. Subtract these two results: $ 2a = 4 \Rightarrow a = 2 $. Substitute back: $ 3(2) + b = 7 \Rightarrow b = 1 $. Then $ 2 + 1 + c = 4 \Rightarrow c = 1 $. Thus, $ p(x) = 2x^2 + x + 1 $.

📌 oxed{2x^2 + x + 1}

📌 Question: Define $ L(u) = u - rac{u^3}{6} $ for all real $ u $. If $ n $ is a positive integer, define $ a_n $ by $ a_1 = rac{\pi}{2} $, $ a_{n+1} = L(a_n) $. Find $ \lim_{n o \infty} a_n $.

📌 Solution: The recurrence $ a_{n+1} = a_n - rac{a_n^3}{6} $ resembles the Taylor series for $ rctan(u) $, where $ rac{d}{du} rctan(u) = rac{1}{1 + u^2} $. However, the recurrence is not exact. Assume the limit $ L $ exists. Then $ L = L - rac{L^3}{6} \Rightarrow rac{L^3}{6} = 0 \Rightarrow L = 0 $. To confirm convergence, note $ a_1 = \pi/2 pprox 1.57 > 1 $, and $ a_{n+1} = a_n(1 - rac{a_n^2}{6}) $. Since $ a_1 < \sqrt{6} $, $ a_n $ is decreasing and bounded below by 0. By monotone convergence, $ a_n o 0 $.

📌 Question: Find the center of the hyperbola $ 4x^2 - 12x - 9y^2 + 18y = 27 $.

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Compute the sum:
\frac{1}{2} \left( \frac{3}{2} - \frac{1}{51} - \frac{1}{52} \right).
Find a common denominator for the subtracted terms:
\frac{1}{51} + \frac{1}{52} = \frac{52 + 51}{51 \cdot 52} = \frac{103}...
\boxed{\frac{3875}{5304}}
Question: Define $ M(v) = v - \frac{v^4}{4} $ for all real $ v $. If $...
b_k = M\left( \frac{1}{k} \right) + M\left( \frac{2}{k} \right) + \cdo...
Find $ b_k $ in simplest form.
M(v) = v - \frac{v^4}{4},
b_k = \sum_{j=1}^{k} M\left( \frac{j}{k} \right) = \sum_{j=1}^{k} \lef...

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